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Sindrei [870]
3 years ago
10

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

re. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 13.0 V. You cut off a 20.0 m length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.60 A. You then cut off a 40.0 m length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.50 A. Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?
Physics
1 answer:
Lostsunrise [7]3 years ago
3 0

Answer:

0.06\Omega/m

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (I=\frac{V}{R}). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

\frac{13}{20z+r} =7.6   .......(1)

\frac{13}{40z+r} =4.5     ......(2)

Substituting the value of r from (2) in (1), we have,

13=152z+7.6\times\frac{13-180z}{4.5}

which simplifying gives us, z=0.0589\Omega/m\approx0.06\Omega/m (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 \Omega

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Answer:

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Explanation:

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3 years ago
In a RLC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o
Marrrta [24]

Answer:

<h2>Case i) if \omega L > \frac{1}{\omega c}</h2><h2>So initially if the circuit is inductive in nature then its net impedance will decrease after this</h2><h2>Case ii) if \omega L < \frac{1}{\omega c}</h2><h2>So initially if the circuit is capacitive in nature then its net impedance will increase after this</h2>

Explanation:

As we know that the impedance of the circuit is given as

z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}

when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination

c_{eq} = c_1 + c_2

so it is

c_{eq} = 2c

now we have

z = \sqrt{(\omega L - \frac{1}{2\omega c})^2 + R^2}

Case i) if \omega L > \frac{1}{\omega c}

So initially if the circuit is inductive in nature then its net impedance will decrease after this

Case ii) if \omega L < \frac{1}{\omega c}

So initially if the circuit is capacitive in nature then its net impedance will increase after this

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3 years ago
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B Ik it all just know who they’ll you
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2 years ago
Bohr’s atomic model differed from Rutherford's because it explained that electrons exist in specified energy levels surrounding
insens350 [35]

Answer:

electrons exist in specified energy levels

Explanation:

In its gold-foil scattering with alpha particles, Rutherford proved that the plum-pudding model of the atom theorised by Thomson was wrong.

From his experiment, Rutherford inferred that the atom actually consists of a very small nucleus, where all the positive charge is concentrated, and the rest of the atom is basically empty, with the electrons (negatively charged) orbiting around the nucleus at very large distance.

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3 years ago
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4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.
kolbaska11 [484]

Answer:

\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg

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